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The drawn component $L_1$ is an unknot. Its exterior is a solid torus $V$, in which $L_2$ is a pattern of a satellite knot. A meridian of a solid torus of $V$ is ${\ell }_1$, whereas its longitudinal direction is $m_1$. Thus the prescribed gluing sends these directions to $m$ and $\ell$, respectively. This identifies $V$ with a tubular neighborhood of $K$ with its zero Seifert framing.

Removing the pattern before making this identification givesEquivalently, meridionally filling the remaining boundary first removes $L_2$ from the construction and leaves $E(K)\cup V=S^3$; its filling core is the required satellite knot $C(K)$.

In the original link diagram, the two parallel passages through a meridional disk of $V$ run in the same direction; closing them uses a single interchange. The winding number of a satellite pattern is therefore two. Untwisting the surrounding disk shows that $P(U)$ is the unknot (the pattern is the $(2,1)$ cable knot, up to the harmless sign of its single interchange). In particular ${\Delta }_{P(U)}=1$. The Satellite formula for the Alexander polynomial now gives the concise answerHere $\doteq$ allows multiplication by $\pm t^j$. Reversing the winding orientation replaces $t^2$ by $t^{-2}$, giving the same Alexander polynomial of a knot up to such a unit.

Let $V=S^1\times D^2$ contain the pattern of a satellite knot $P$. Its class in $H_1(V;\mathbb{Z})\cong \mathbb{Z}$ is the winding number of a satellite pattern. Winding number zero therefore makes $P$ null-homologous in $V$, so $P$ bounds an oriented embedded surface $F_P\subset V$. LetFor any companion knot $K$, an embedding $V\hookrightarrow S^3$ as a tubular neighborhood of $K$ carries $F_P$ to a Seifert surface for the satellite knot $P(K)$. Henceand the bound depends only on the pattern.